Let A¯=i¯−3j→+4k¯,B¯=6i¯+4j→−8k→,C¯=5i¯+2j¯+5k¯ and a vector R¯ satisfies R¯×B¯=C¯×B¯,R¯⋅A¯=0 then value of |B¯||R¯−C¯| is

R¯×B¯=C¯×B¯⇒A¯×(R¯×B¯)=A¯×(C¯×B¯)⇒A¯.B¯R¯-A¯.R¯B¯=A¯.B¯C¯-A¯.C¯B¯⇒A¯.C¯B¯=A¯.B¯C¯-A¯.B¯R¯=A¯.B¯C¯-R¯ (∵A¯.R¯=0)∴B¯C¯-R¯=A¯.B¯A¯.C¯=3819=2

Let A¯=i¯−3j→+4k¯,B¯=6i¯+4j→−8k→,C¯=5i¯+2j¯+5k¯ and a vector R¯ satisfies R¯×B¯=C¯×B¯,R¯⋅A¯=0 then value of |B¯||R¯−C¯| is

R¯×B¯=C¯×B¯⇒A¯×(R¯×B¯)=A¯×(C¯×B¯)⇒A¯.B¯R¯-A¯.R¯B¯=A¯.B¯C¯-A¯.C¯B¯⇒A¯.C¯B¯=A¯.B¯C¯-A¯.B¯R¯=A¯.B¯C¯-R¯ (∵A¯.R¯=0)∴B¯C¯-R¯=A¯.B¯A¯.C¯=3819=2

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$Let \bar{A} = \bar{i} - 3 \vec{j} + 4 \bar{k}, \bar{B} = 6 \bar{i} + 4 \vec{j} - 8 \vec{k}, \bar{C} = 5 \bar{i} + 2 \bar{j} + 5 \bar{k} and a vector \bar{R} satisfies$ $\bar{R} \times \bar{B} = \bar{C} \times \bar{B}, \bar{R} \cdot \bar{A} = 0 then value of \frac{| \bar{B} |}{| \bar{R} - \bar{C} |} is$

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Detailed Solution

$\begin{array}{l} \bar{R} \times \bar{B} = \bar{C} \times \bar{B} \\ \Rightarrow \bar{A} \times (\bar{R} \times \bar{B}) = \bar{A} \times (\bar{C} \times \bar{B}) \\ \Rightarrow (\bar{A} . \bar{B}) \bar{R} - (\bar{A} . \bar{R}) \bar{B} = (\bar{A} . \bar{B}) \bar{C} - (\bar{A} . \bar{C}) \bar{B} \\ \Rightarrow (\bar{A} . \bar{C}) \bar{B} = (\bar{A} . \bar{B}) \bar{C} - (\bar{A} . \bar{B}) \bar{R} = (\bar{A} . \bar{B}) (\bar{C} - \bar{R}) (∵ \bar{A} . \bar{R} = 0) \\ ∴ \frac{|\bar{B}|}{|\bar{C} - \bar{R}|} = \frac{|\bar{A} . \bar{B}|}{|\bar{A} . \bar{C}|} = \frac{38}{19} = 2 \end{array}$